LeetCode——Remove Duplicates from Sorted Array

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本文介绍了一种在原地去除排序数组中重复元素的方法，通过两种不同的算法实现：使用HashMap和双指针技巧。双指针技巧尤其高效，运行速度达到100%的Java在线提交中最快。文章提供了详细的代码示例和性能分析。

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Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:
Given nums = [1,1,2],
Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn’t matter what you leave beyond the returned length.
Example 2:
Given nums = [0,0,1,1,1,2,2,3,3,4],
Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
It doesn’t matter what values are set beyond the returned length.
Clarification:

Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

解法一——利用HashMap

其实完全用不着HashMap，只是一开始脑子没转过弯来，用HashMap的话在检查map中是否已存在此整数时会比较费时间

public int removeDuplicates(int[] nums) 
{
	int length=0;
	HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
	for(int i=0;i<nums.length;i++)
	{
		if(!map.containsKey(nums[i]))
		{	
            nums[length]=nums[i];
			map.put(nums[i],1);
			length++;
		}
	}
	return length;
 }

Runtime: 8 ms, faster than 45.45% of Java online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 42.1 MB, less than 35.04% of Java online submissions for Remove Duplicates from Sorted Array.

解法二——利用两个索引

合理利用length 用其对比数组中的数字是否和nums[length]相等，不相等则更新length ，最后要注意length指向的值

public int removeDuplicates(int[] nums) 
	{
		if(nums==null||nums.length==0)
			return 0;
		int length=0;
		for(int i=1;i<nums.length;i++)
		{
			if(nums[i]!=nums[length])
			{
				length++;
				nums[length]=nums[i];
			}
		}
        length++;
		return length;
	 }

Runtime: 5 ms, faster than 100.00% of Java online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 42.2 MB, less than 32.08% of Java online submissions for Remove Duplicates from Sorted Array.